What is the limit of the function f(x) = sin(x)/x as x approaches zero?

What is the limit of the function f(x) = sin(x)/x as x approaches zero?

• A. 1
• B. 0
• C. Infinity
• D. -1

Answer: (A)

To find the limit of the function \( f(x) = \frac{\sin(x)}{x} \) as \( x \) approaches zero, we can utilize the well-known limit property in calculus. This limit can be analyzed through various methods, including using L'Hôpital's Rule, the Squeeze Theorem, or recognizing the Taylor series expansion of \( \sin(x) \). When we approach this limit directly by substituting \( x = 0 \), we encounter an indeterminate form \( \frac{0}{0} \). However, from calculus, we know that as \( x \) gets very close to zero, the value of \( \sin(x) \) behaves similarly to \( x \). More specifically, the Taylor series expansion for \( \sin(x) \) around zero is: \[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots \] When substituting \( \sin(x) \) into the limit expression we have: \[ \lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac

To find the limit of the function ( f(x) = \frac{\sin(x)}{x} ) as ( x ) approaches zero, we can utilize the well-known limit property in calculus. This limit can be analyzed through various methods, including using L'Hôpital's Rule, the Squeeze Theorem, or recognizing the Taylor series expansion of ( \sin(x) ).

When we approach this limit directly by substituting ( x = 0 ), we encounter an indeterminate form ( \frac{0}{0} ). However, from calculus, we know that as ( x ) gets very close to zero, the value of ( \sin(x) ) behaves similarly to ( x ). More specifically, the Taylor series expansion for ( \sin(x) ) around zero is:

[

\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots

]

When substituting ( \sin(x) ) into the limit expression we have:

[

\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac